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21 tháng 8 2016

\(S=-1^2+2^2-3^2+4^2-...+2016^2\)

\(=\left(2-1\right)\left(2+1\right)+\left(4-3\right)\left(4+3\right)+...+\left(2016-2015\right)\left(2016+2015\right)\)

\(=3+7+..+4031\)

\(=2033136\)

\(A=\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-\frac{1}{15}\times4^{64}\)

\(15A=\left(4^2-1\right)\left(4^2+1\right)\left(4^4+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-4^{64}\)

\(15A=\left(4^4-1\right)\left(4^4+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-4^{64}\)

\(15A=\left(4^{16}-1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-4^{64}\)

\(15A=\left(4^{32}-1\right)\left(4^{32}+1\right)-4^{64}\left(4^{32}\right)\)

\(15A=4^{64}-1-4^{64}\)

\(A=-\frac{1}{15}\)

 

3: =(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)

=(5^4-1)(5^4+1)(5^8+1)(5^16+1)

=(5^8-1)(5^8+1)(5^16+1)

=(5^16-1)(5^16+1)

=5^32-1

4:

D=(4^4-1)(4^4+1)(4^8+1)*....*(4^64+1)

=(4^8-1)(4^8+1)*...*(4^64+1)

=...

=4^128-1

5: =(5^2-1)(5^2+1)(5^4+1)*...*(5^128+1)+(5^256-1)

=(5^4-1)(5^4+1)*...*(5^128+1)+5^256-1

=5^256-1+5^256-1

=2*5^256-2

7 tháng 7 2023

thsu là rất ngưỡng mộ anh ạ 🥹 em mấy lần off vì quá nhác nhưng lần nào ngoi lại lên cũng thấy anh cày chăm chỉ quá tr 😭

19 tháng 6 2019

\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)

\(\Rightarrow B=2^{64}-1-2^{64}=-1\)

19 tháng 6 2019

a) Đặt \(A=\left(\frac{1}{2}+1\right).\left(\frac{1}{4}+1\right).\left(\frac{1}{16}+1\right)...\left(1+\frac{1}{2^{2n}}\right)\)

Rút gọn:  \(A=\frac{2+1}{2}.\frac{4+1}{4}.\frac{16+1}{16}...\frac{2^{2.n}+1}{2^{2.n}}=\frac{2^{2.0}+1}{2^{2.0}}.\frac{2^{2.1}+1}{2^{2.1}}.\frac{2^{2.2}+1}{2^{2.2}}...\frac{2^{2.n}+1}{2^{2.n}}\)

\(\Rightarrow A=\frac{\left(2^{2.0}+1\right).\left(2^{2.1}+1\right).\left(2^{2.2}+1\right)...\left(2^{2.n}+1\right)}{2^{2.0}.2^{2.1}.2^{2.2}...2^{2.n}}.\)

b) Đặt \(B=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}\)

\(\Leftrightarrow B=\left(2-1\right).\left(2+1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^2-1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(\Leftrightarrow B=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^8-1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}\)

\(\Leftrightarrow B=\left(2^{16}-1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}=\left(2^{32}-1\right).\left(2^{32}+1\right)-2^{64}\)

\(\Leftrightarrow B=2^{64}-1-2^{64}=-1\)Vậy B =-1.

14 tháng 7 2016

a) (2+1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2+1)(2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2^2-1)(2^2+1)(2^4+1)...(2^32+1)-2^64

=(2^4-1)(2^4+1)....(2^32+1)-2^64

=......

=(2^32-1)(2^32+1)-2^64

=2^64-1-2^64=-1

b)Đặt A=(5+3)(5^2+3^2)(5^4+3^4)...(5^64+3^64)+(5^128-3^128)/2

đặt B=(5+3)(5^2+3^2)(5^4+3^4)...(5^64+3^64)

\(2B=\left(5-3\right)\left(5+3\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)

\(2B=\left(5^2-3^2\right)\left(5^2+3^2\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)

\(2B=\left(5^4-3^4\right)\left(5^4+3^4\right)...\left(5^{64}+3^{64}\right)\)

\(2B=.......\)

2B=(5^64-3^64)(5^64+3^64)

2B=5^128-3^128

B=(5^128-3^128)/2 (thế vào đề bài)

=> A=B+(5^128-3^128)/2=(5^128-3^128)/2+(5^128-3^128)/2=\(\frac{2\left(5^{128}-3^{128}\right)}{2}=\left(5^{128}-3^{128}\right)\)

14 tháng 7 2016

a) A = ( 2-1)(2+1)(22+1)...(232+1)-264

         =(22-1)(22+1)(24+1)... -264

       =....

       =264-1-264=1

câu b tương tự nhá

18 tháng 5 2017

\(A=\frac{1}{841}\)

30 tháng 8 2017

làm kiểu j thế

3 tháng 4 2017

Bạn chú ý cách viết phương trình.

Phương trình chỉ có dạng f(x)=g(x) thôi, không có dạng A=f(x)=g(x) như bạn viết.

\(VT=\left[8\left(x+\frac{1}{x}\right)^2-4\left(x^2+\frac{1}{x^2}\right)\left(x+\frac{1}{x}\right)^2\right]+4\left(x^2+\frac{1}{x^2}\right)^2\)

\(=4\left(x+\frac{1}{x}\right)^2\left(2-x^2-\frac{1}{x^2}\right)+4\left(x^2+\frac{1}{x^2}\right)^2\)

\(=-4\left(x+\frac{1}{x}\right)^2\left(x-\frac{1}{x}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2\)

\(=-4\left(x^2-\frac{1}{x^2}\right)^2+4\left(x^2+\frac{1}{x^2}\right)^2\)

\(=-4x^4+8-\frac{4}{x^4}+4x^4+8+\frac{4}{x^4}\)

\(=16\)

Phương trình đã cho trở thành

\(\left(x+4\right)^2=16\\ \Leftrightarrow\orbr{\begin{cases}x+4=-4\\x+4=4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-8\\x=0\end{cases}}\)

25 tháng 5 2018

Phải là (2+1)(2²+1)(2⁴+1)...(2³²+1)- 2^64

(2+1)(2²+1)(2⁴+1)...(2³²+1)

=(2-1)(2+1)(2²+1)(2⁴+1)...(2³²+1)

=(2²-1)(2²+1)(2⁴+1)...(2³²+1)

=(2⁴-1)(2⁴+1)...(2³²+1)=…=2^64-1

Vậy C=-1

23 tháng 8 2015

3  = 2^2 - 1 

Áp dụng HĐT a^2 - b^2 

kq : 2^128 - 1 

24 tháng 7 2016

đặt 3(42 + 1 )(4+1)......=A

5A=15(42+1)(44+1)......

=(42-1)(42+1)(44+1).....

=(44-1)(44+1).......

bạn giái tiếp là ra

24 tháng 7 2016

= 3[(42.44.48.416.432)+(1.1.1.1.1)]

= 3[432768 + 1]

= 1232768 + 3

Ko chắc nhé